$a,A=\frac{-1}{3}+\frac{-1}{3^{3}}+...+\frac{-1}{3^{101}} $

$\Rightarrow 9A=\frac{-9}{3}+\frac{-9}{3^{3}}+\frac{-9}{3^{5}}...+\frac{-9}{3^{101}} $

$=-3+\frac{-1}{3}+\frac{-1}{3^{3}}...+\frac{-1}{3^{99}} $

$\Rightarrow 9A-A=(-3+\frac{-1}{3}+\frac{-1}{3^{3}}...+\frac{-1}{3^{99}})-(\frac{-1}{3}+\frac{-1}{3^{3}}+...+\frac{-1}{3^{101}}) $

 $\Rightarrow 8A=-3+\frac{1}{3^{101}}=\frac{-3^{102}+1}{3^{101}}$  $\Rightarrow A=\frac{-3^{102}+1}{8.3^{101}}$

$b,\frac{x+1}{2015}+\frac{x+3}{671}+\frac{x+6}{402}+9=0 $

$\Leftrightarrow (\frac{x+1}{2015}+1)+(\frac{x+3}{671}+3)+(\frac{x+6}{402}+5)=0 $

$\Leftrightarrow\frac{x+2016}{2015}+\frac{x+2016}{671}+\frac{x+2016}{402}=0 $

$\Leftrightarrow (x+2016)(\frac{1}{2015}+\frac{1}{671}+\frac{1}{402})=0 $

$\Leftrightarrow x+2016=0 $$\Leftrightarrow x=-2016 $

Vậy PT có nghiệm duy nhất $ x=-2016 $

$c,A=\frac{15}{4}+\frac{30}{9}+...+\frac{30003}{10000} $

$=(3+\frac{3}{4})+(3+\frac{3}{9})+...+(3+\frac{3}{10000}) $

$=(3+\frac{3}{2^2})+(3+\frac{3}{3^2})+...+(3+\frac{3}{100^2}) $

$=297+3(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{100^2}) $

$Có:\ \frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{100^2}>\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{100.101}=\frac{3-2}{2.3}+\frac{4-3}{3.4}+...+\frac{101-100}{100.101} =  \frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{100}-\frac{1}{101}=\frac{1}{2}-\frac{1}{101}>\frac{1}{2}-\frac{1}{6}=\frac{1}{3}$

$Tương\ tự:\ \frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{100^2}<\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}=1-\frac{1}{100}<1$ 

Suy ra:$297+3.\frac{1}{3}<A<297+3.1$ 
$\Leftrightarrow 298<A<300$(đpcm)

b)

Ta có :

$(\dfrac{x+1}{2015}+1)+(\dfrac{x+3}{671}+3)+(\dfrac{x+6}{402}+5)=0$

$⇔ \dfrac{x+2016}{2015}+\dfrac{x+2016}{671}+\dfrac{x+2016}{402}=0$

$⇔x+2016=0$

$⇔x=-2016$